Maclaurin Partial Fractions C10029

1. **State the problem:** We want to find the MacLaurin series coefficients of the function $$F(x) = \frac{x}{1 - x - x^2}$$ using partial fractions within its region of convergence. 2. **Find the roots of the denominator:** Solve $$1 - x - x^2 = 0$$ for $x$. 3. Using the quadratic formula, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a = -1$, $b = -1$, $c = 1$, we get: $$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$$ 4. Let $$r_1 = \frac{1 + \sqrt{5}}{2}$$ and $$r_2 = \frac{1 - \sqrt{5}}{2}$$ be the roots. 5. **Express the denominator as:** $$1 - x - x^2 = -(x - r_1)(x - r_2)$$ 6. **Rewrite the function:** $$F(x) = \frac{x}{1 - x - x^2} = \frac{x}{-(x - r_1)(x - r_2)} = -\frac{x}{(x - r_1)(x - r_2)}$$ 7. **Set up partial fractions:** $$-\frac{x}{(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$ 8. Multiply both sides by $(x - r_1)(x - r_2)$: $$-x = A(x - r_2) + B(x - r_1)$$ 9. Expand: $$-x = A x - A r_2 + B x - B r_1 = (A + B) x - (A r_2 + B r_1)$$ 10. Equate coefficients: - Coefficient of $x$: $$-1 = A + B$$ - Constant term: $$0 = - (A r_2 + B r_1) \Rightarrow A r_2 + B r_1 = 0$$ 11. From $A + B = -1$, we get $$B = -1 - A$$. 12. Substitute into $A r_2 + B r_1 = 0$: $$A r_2 + (-1 - A) r_1 = 0 \Rightarrow A (r_2 - r_1) = r_1$$ 13. Solve for $A$: $$A = \frac{r_1}{r_2 - r_1}$$ 14. Then $$B = -1 - A = -1 - \frac{r_1}{r_2 - r_1} = \frac{r_2 - r_1 - r_1}{r_2 - r_1} = \frac{r_2 - 2 r_1}{r_2 - r_1}$$ 15. **Simplify denominators:** Note that $$r_2 - r_1 = \frac{1 - \sqrt{5}}{2} - \frac{1 + \sqrt{5}}{2} = -\sqrt{5}$$ 16. So, $$A = \frac{r_1}{-\sqrt{5}} = -\frac{r_1}{\sqrt{5}}$$ $$B = \frac{r_2 - 2 r_1}{-\sqrt{5}} = -\frac{r_2 - 2 r_1}{\sqrt{5}}$$ 17. **Rewrite $F(x)$:** $$F(x) = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$ 18. **Express each term as a geometric series:** For $|x| < |r_1|$ and $|x| < |r_2|$, $$\frac{1}{x - r} = -\frac{1}{r} \cdot \frac{1}{1 - \frac{x}{r}} = -\frac{1}{r} \sum_{n=0}^\infty \left(\frac{x}{r}\right)^n = -\sum_{n=0}^\infty \frac{x^n}{r^{n+1}}$$ 19. Therefore, $$F(x) = A \left(-\sum_{n=0}^\infty \frac{x^n}{r_1^{n+1}}\right) + B \left(-\sum_{n=0}^\infty \frac{x^n}{r_2^{n+1}}\right) = -\sum_{n=0}^\infty \left( \frac{A}{r_1^{n+1}} + \frac{B}{r_2^{n+1}} \right) x^n$$ 20. **MacLaurin series coefficients:** $$c_n = - \left( \frac{A}{r_1^{n+1}} + \frac{B}{r_2^{n+1}} \right)$$ 21. **Region of convergence:** The series converges for $$|x| < \min(|r_1|, |r_2|) = |r_2|$$ since $|r_2| < |r_1|$. **Final answer:** The MacLaurin series of $$F(x)$$ is $$F(x) = \sum_{n=0}^\infty c_n x^n$$ with $$c_n = - \left( \frac{A}{r_1^{n+1}} + \frac{B}{r_2^{n+1}} \right)$$ where $$A = -\frac{r_1}{\sqrt{5}}, \quad B = -\frac{r_2 - 2 r_1}{\sqrt{5}}, \quad r_1 = \frac{1 + \sqrt{5}}{2}, \quad r_2 = \frac{1 - \sqrt{5}}{2}$$ and the series converges for $$|x| < |r_2|$$.